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October 15

Sum of cubes of Pythagorean triples

A recent question on WP:RD/M suggests that (3,4,5) is the only primitive Pythagorean triple whose cubes sum to a prefect cube. Is this true? -- 110.49.12.136 (talk) 01:03, 15 October 2011 (UTC)[reply]

Direct inspection of all primitives (a,b,c) with c < 2 billion shows no other examples, which makes this very plausible. Dragons flight (talk) 03:36, 15 October 2011 (UTC)[reply]

Define your entities! For even exponents, employing negative bases yields no essentially new solutions. In this case, however, if negative integers are allowed, we get further solutions. Recall the Ramanujan number 1729 = 1³+12³ = 9³+10³; from this you get e.g. the triple (-1,9,10). JoergenB (talk) 14:19, 16 October 2011 (UTC)[reply]

I've heard that referred to as Ramanujan's taxi, but wasn't it really Hardy's taxi? In any case, (-1,9,10) isn't a Pythagorean triple, and if we open to all triples, we can include many others such as (1,6,8). -- 49.229.147.145 (talk) 16:31, 16 October 2011 (UTC)[reply]

Alice and Bob in the Candy Shop

Hi. Alice is in a candy shop with five different kinds of candy, and she needs to purchase treats to make goody bags for her two sons. How many ways can she purchase such that between the two bags every kind is represented , but there are only (=exactly) two kinds that both bags have? (ignore who gets what bag and how many individual pieces of candy are in each bag).

I attempted to solve by combinatorics: I reasoned that either one bag must contain 3 types of candy and the other 4, or one contains 2 and the other 5. By choosing the types that go into the first bag those that must go into the second bag are determined also, so the answer should be the some of ways there are to choose 3 of the 5 kinds and 2 of 5, thus: ${5 \choose 3}+{5 \choose 2}=20$ . Is this right? Thanks — Preceding unsigned comment added by 118.96.49.118 (talk) 03:50, 15 October 2011 (UTC)[reply]

No, because in the 3 and 4 case, choosing the types that go into the first bag doesn't determine the types that go into the second bag; you also need to choose which 2 are the repeated 2. So for the 3 in one, 4 in the other case, there are

{5 \choose 3}\cdot {3 \choose 2}

ways. There's also a bit of ambiguity over whether the two goody bags are distinguishable. If I swap them, does that count as a different way? You've assumed it doesn't, which seems fine.--121.74.125.249 (talk) 05:21, 15 October 2011 (UTC)[reply]

If you really want to know how many ways she can make the purchase, the answer is 10, because all she needs to decide is which two types of candy to get two of, and there are ${5 \choose 2}$ ways of doing that. But if you want to know how many ways there are to split the candy into two bags, the answer is 80 -- for each of the 10 possible purchases, there are 2³ = 8 ways of assigning the types of which only one has been purchased to bags. Looie496 (talk) 05:34, 15 October 2011 (UTC)[reply]

Who is the Bob from the title? – b_jonas 16:54, 17 October 2011 (UTC)[reply]

Circle probability

Good evening. I have a problem that states: "Two cricles of radius 1 are constructed as follows: The centre (x,0) of circle A is chosen at random from the interval [0,2] and the centre (1,y) of circle B from the segment joining ~~(0,1) and (2,1)~~ CORRECTION: (1,0) and (1,2). What is the probability the circles intersect?" My thinking and working: The circles intersect iff the distance between their centres is less than or equal to 2. Therefore I am looking as an intermediary step for the set of ordered pairs given by $\{(x,y)|{\sqrt {(x-1)^{2}+y^{2}}}\leq 2,y\geq 0\}$ , equivalently the area under a semicircle which I calculate to be 2π. The area of the closed rectangle [0,2]×[0,2] is 4 so the probability should be π/2. HOWEVER I am not supposed to be turning up pi in my answer because the choices given are $A.{\frac {2+{\sqrt {2}}}{2}},B{\frac {3{\sqrt {3}}+2}{8}},C{\frac {2{\sqrt {2}}-1}{2}},D{\frac {2+{\sqrt {3}}}{4}}orE{\frac {4{\sqrt {3}}-3}{4}}$ Please help me to see where my working is faulty (a measure theory perspectived approach is also appreciated). Thank you. 80.78.66.116 (talk) 04:43, 15 October 2011 (UTC)[reply]

I assume that you meant to write "the centre (1,y) of circle B from the segment joining ~~(0,1) and (2,1)~~ (1,0) and (1,2)." -- 110.49.12.136 (talk) 05:24, 15 October 2011 (UTC)[reply]

Actually, I assume he or she meant the other way around: "the centre ~~(1,y)~~ (y,1) of circle B from the segment joining (0,1) and (2,1)", since that version gives one of the listed answers, and the other doesn't.

That would make sense, as I wasn't getting one of the listed answers my way either. -- 110.49.12.136 (talk) 06:13, 15 October 2011 (UTC)[reply]

To the original poster, your set of ordered pairs is wrong, because you've made a mistake applying the distance formula.--121.74.125.249 (talk) 05:41, 15 October 2011 (UTC)[reply]

~~As a hint, over what values does x range in your radius 2 semicircle, and over what values is x supposed to range?~~ -- 110.49.12.136 (talk) 05:32, 15 October 2011 (UTC) This hint applies to a misinterpretation of the problem. -- 110.49.12.136 (talk) 11:46, 15 October 2011 (UTC)[reply]

I have made the correction. 110's first interpretation was correct. It is possible that correct answer is missing; that's why I am asking about it here. 80.78.66.116 (talk) 14:32, 15 October 2011 (UTC)[reply]

I did the problem and got one of the answers on the list so I think you're still doing something wrong. The difference in vertical coordinates is constant, and the real question is what is the probability that the two centers will be distance at most 2 from each other. So then the question reduces the probability the x and y differ by at most √3. It helped to sketch this out as a region in the x-y plane and find its area as a proportion of the area of a 2 by 2 square.--RDBury (talk) 15:19, 15 October 2011 (UTC)[reply]

The 2 by 2 square loses identical isosceles triangles at opposite corners (they define cases when the circles do not overlap), and the proportionate remaining area is that given in Answer E. There were actually only four sensible possibilities, as Answer A is greater then 1.←86.155.185.195 (talk) 17:55, 15 October 2011 (UTC)[reply]

If the way you've written it now is the way it's written in the book (or wherever you're getting this from), then the book has a typo. The problem has B varying vertically, but the answers assume it varies horizontally.--121.74.125.249 (talk) 20:25, 15 October 2011 (UTC)[reply]

IP 110 here. I fully agree with 121. I suggested the correction based simply on the inconsistency, before I worked the problem, but when I did, I got answers with π -- more complicated than yours due to the hint I struck out above (which required an integral), but not one of the provided answers. Working the problem as interpreted by 121 with "centre (y,1) of circle B" I do get one of the provided answers. If the book gives the problem as you've written it now, then perhaps the grad student who worked out the answers for his prof writing the book misinterpreted the problem. -- 110.49.27.51 (talk) 01:00, 16 October 2011 (UTC)[reply]

error in a measurement

(This might belong on the Science desk, I'm not sure). If I have some measurement x with error e, and I have to calculate some quantity exp(x), what is the error of exp(x)? --121.98.230.200 (talk) 08:23, 15 October 2011 (UTC)[reply]

exp(x+e)=exp(e)exp(x), so the absolute error exp(x+e)-exp(x) is exp(x)[exp(e)-1], the relative error being exp(e)-1.86.155.185.195 (talk) 13:00, 15 October 2011 (UTC)[reply]

If e is small you can use e as an estimate for exp(e)-1. So the error is e exp(x).--RDBury (talk) 15:41, 15 October 2011 (UTC)[reply]

1+1=2

for example, an alien ask you how to prove 1+1=2. how would you answer? — Preceding unsigned comment added by 203.112.82.1 (talk) 13:40, 15 October 2011 (UTC)[reply]

It depends on how that alien would define the concept of "2". Human mathematicians essentially define 2 as 1+1 (strictly speaking, the successor of 1), so there isn't much to show. Sławomir Biały (talk) 14:40, 15 October 2011 (UTC)[reply]

Indeed, all such proofs will rely upon some sort of axioms. The Peano axioms are commonly used to prove this sort of thing in undergrad math classes. For a more detailed construction/ proof, see Natural_number#Constructions_based_on_set_theory. If the aliens believe in the notion of a set, and set inclusion, they will be forced to agree that 1+1=2 :) SemanticMantis (talk) 14:46, 15 October 2011 (UTC)[reply]

The interesting problem is why is it a constant? If you stick two rabbits together the number may not stay 2 for very long. And how about clouds? And when you put an apple and an orange together you don't get either two apples or two oranges, yu get something different again - fruit. Once you've got past that it is 2 by definition as that constant. ;-) Dmcq (talk) 14:49, 15 October 2011 (UTC)[reply]

Another interesting problem concerns the meaning of addition. A hypothetical dolphin mathematician might define "1" and "2" in terms of acoustic wave forms. Addition of natural numbers may not even be well-defined (or easily defined) for them, but addition (superposition) of waves would be. A natural axiomatic system for them might look very strange to us. Sławomir Biały (talk) 15:09, 15 October 2011 (UTC)[reply]

And to mangle an old nerd joke about binary numbers, there are 10 kinds of people in the world.... HiLo48 (talk) 01:34, 16 October 2011 (UTC)[reply]

"There are 10 kinds of people in the world: those who go about dividing the world into 10 kinds of people, and those who don't." :) -- Jack of Oz ^{[your turn]} 09:34, 18 October 2011 (UTC)[reply]

File:Principia Mathematica theorem 54-43.png

RDBury's plan for saving Earth from alien invasion ... in our time.

Shouldn't the binary form of the joke be "There are 11 kinds of people in the world: those that can count, and those that can't." — Fly by Night (talk) 17:38, 21 October 2011 (UTC)[reply]

Principia Mathematica famously has a proof that 1+1=2, so you could hand the aliens that and tell them to come back when they've understood it. That should keep them out of our hair for a few lifetimes.--RDBury (talk) 15:27, 15 October 2011 (UTC)[reply]

Closure of C

I've noticed that for the structures up to R you can "break out" (defining "breaking out" of a structure as performing some (binary?) operation so that the result is not within the structure): examples: you can "break out" of N by subtracting a bigger element from a smaller element, you can "break out" of Z by dividing elements that don't divide cleanly, you can break out of Q⁺ by taking square roots, and you can break out of R by taking even roots of negative numbers. In more proper terminology N is not closed under subtraction, nor Z under division, nor Q nor R under fractional exponents. Is there any way to "break out" of C, and if so what is the most elementary operation that can be used to do it? Thanks. 24.92.85.35 (talk) 18:18, 15 October 2011 (UTC)[reply]

Extension of algebraic structures is usually thought of in terms of equations which are insolvable in the original structure -

x+1=0

has no solution in N etc. In C there's no solution to

(ix-xi)^{2}=-4

- adding a solution gives rise to the quaternions. -- Meni Rosenfeld (talk) 18:48, 15 October 2011 (UTC)[reply]

C is usually thought of as the last in the series of extensions you get by asking questions like "what is the answer to x² + 1 = 0?", so in a sense the answer is there is none. But the above is one way (which I've not seen before), another is considering x + 1 = x which is solved by making x infinite, so you extend C (or R) with infinity. This is one way to define a projective space, although that's usually done for geometric reasons not algebraic ones.--JohnBlackburne^words_deeds 19:41, 15 October 2011 (UTC)[reply]

Just a note on terminology to add to the previous answers: C is an algebraically closed field. AndrewWTaylor (talk) 09:58, 18 October 2011 (UTC)[reply]